1. Cho a,b>0; a+b=1
Tìm min A=\(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\)
2. Cho x,y,x >0 t/m: \(x^2+y^2+z^2=3\)
CMR: \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\) ≥ 3
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NH
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 1 2019
\(\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)
\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\ge\dfrac{2\sqrt{bc}}{a}.\dfrac{2\sqrt{ac}}{b}.\dfrac{2\sqrt{ab}}{c}=8\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
TT
1
C
22 tháng 9 2019
Áp dụng BĐT Cô -si cho 3 số dương:
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
NB
1
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2019
a/ Ta có \(\dfrac{\left(a+b\right)^2}{4}\ge ab\Rightarrow\left(a+b\right)^2\ge4\Rightarrow a+b\ge2\)
\(\left(a+1\right)\left(b+1\right)=ab+\left(a+b\right)+1=a+b+2\ge2+2=4\) (đpcm)
Dấu "=" xảy ra khi \(a=b=1\)
b/ Áp dụng BĐT \(ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow ab\le\dfrac{1}{4}\Rightarrow\dfrac{1}{ab}\ge4\)
Lại áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\) cho 2 số dương ta được:\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=\dfrac{1}{2}\left(1+\dfrac{1}{ab}\right)^2\ge\dfrac{1}{2}\left(1+4\right)^2=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
NH
1
HN
20 tháng 4 2017
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right)\dfrac{9}{a+b+c}=9\)
PC
21 tháng 4 2016
=1+1/a+1/b+1/ab (1)
Áp dụng Cosy ta có 1/a+1/b>=4/(a+b)=4 (2)
(a+b)^2>=4ab nên ab<=(a+b)^2/4=1/4 hay 1/ab>=4 (3)
Từ (1)(2)(3) ta đc 1+1/a+1/b+1/ab>=1+4+4=9 (đpcm)
BC
21 tháng 4 2016
Ta có: \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\) \(=\left(1+1+\frac{b}{a}\right)\left(1+1+\frac{a}{b}\right)\) \(=\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\) \(=4+2\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{ab}{ab}\) \(=5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)
. Áp dụng BĐT Cô-si cho 2 số \(\frac{a}{b}\) và \(\frac{b}{a}\) , ta có:
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ab}}=2\) . Suy ra \(2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4\)
. Suy ra \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge5+4=9\) (đpcm)
. Dấu "=" xảy ra khi \(a=b\)
\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Dấu \("="\Leftrightarrow x=y\)
\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)
\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)
Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)
Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)
Dấu \("="\Leftrightarrow x=y=z=1\)